Problem: Recently, Anton has found a set. The set consists of small English letters. Anton carefully wrote out all the letters from the set in one line, separated by a comma. He also added an opening curved bracket at the beginning of the line and a closing curved bracket at the end of the line. 

Unfortunately, from time to time Anton would forget writing some letter and write it again. He asks you to count the total number of distinct letters in his set.

Input

The first and the single line contains the set of letters. The length of the line doesn't exceed 1000. It is guaranteed that the line starts from an opening curved bracket and ends with a closing curved bracket. Between them, small English letters are listed, separated by a comma. Each comma is followed by a space.

Output

Print a single number — the number of distinct letters in Anton's set.

Examples

Input

{a, b, c}


Output

3


Input

{b, a, b, a}


Output

2


Input

{}


Output

0

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long double pi = acos(-1.0);
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
bool checkprime(long long n) {
  long long i = 2;
  while (i * i <= n) {
    if (n % i == 0) return 0;
    i++;
  }
  return 1;
}
bool check(long long a[], long long n) {
  for (long long i = 0; i < n; ++i)
    if (a[i] != i + 1) return 1;
  return 0;
}
long long fact(long long n) {
  if (n == 0) return 1;
  return n * fact(n - 1);
}
long long sod(long long n) {
  long long sum = 0;
  while (n) {
    sum += n % 10;
    n /= 10;
  }
  return sum;
}
vector<long long> v;
long long bs(long long low, long long high, long long val) {
  long long mid = (low + high) / 2;
  if (low == high) return high;
  if (v[mid] >= val)
    return bs(low, mid, val);
  else
    return bs(mid + 1, high, val);
}
void lucky(long long n, long long cur, long long ct4, long long ct7,
           long long val) {
  if (n == cur) {
    if (ct4 == ct7) {
      v.push_back(val);
    }
    return;
  }
  lucky(n, cur + 1, ct4 + 1, ct7, val * 10 + 4);
  lucky(n, cur + 1, ct4, ct7 + 1, val * 10 + 7);
}
void DJ() {
  set<char> a;
  string s;
  getline(cin, s);
  for (long long i = 0; i < s.size(); ++i) {
    if ((s[i] - 'a') >= 0 && (s[i] - 'a') <= 25) a.insert(s[i]);
  }
  cout << (long long)a.size();
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t = 1;
  while (t--) DJ();
  return 0;
}